Problem: What is the value of $\dfrac{d}{dx}\left(\dfrac{2x+3}{3x^2-4}\right)$ at $x=-1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4$ (Choice B) B $-1$ (Choice C) C $\dfrac14$ (Choice D) D $-32$
Solution: Let's first find the expression for $\dfrac{d}{dx}\left(\dfrac{2x+3}{3x^2-4}\right)$ (i.e. for any input value $x$ ). Then, we can plug $x=-1$ and evaluate. $\dfrac{2x+3}{3x^2-4}$ is a rational expression. To find the derivative of rational expressions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d d x ( 2 x + 3 3 x 2 − 4 ) = ( 3 x 2 − 4 ) d d x ( 2 x + 3 ) − ( 2 x + 3 ) d d x ( 3 x 2 − 4 ) ( 3 x 2 − 4 ) 2 = ( 3 x 2 − 4 ) ( 2 ) − ( 2 x + 3 ) ( 6 x ) ( 3 x 2 − 4 ) 2 = 6 x 2 − 8 − 12 x 2 − 18 x ( 3 x 2 − 4 ) 2 = − 6 x 2 − 18 x − 8 ( 3 x 2 − 4 ) 2 The quotient rule Differentiate ( 2 x + 3 ) & ( 3 x 2 − 4 ) Expand \begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac{2x+3}{3x^2-4}\right) \\\\ &=\dfrac{(3x^2-4)\dfrac{d}{dx}(2x+3)-(2x+3)\dfrac{d}{dx}(3x^2-4)}{(3x^2-4)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{(3x^2-4)(2)-(2x+3)(6x)}{(3x^2-4)^2}&&\gray{\text{Differentiate }(2x+3)\text{ & }(3x^2-4)} \\\\ &=\dfrac{6x^2-8-12x^2-18x}{(3x^2-4)^2}&&\gray{\text{Expand}} \\\\ &=\dfrac{-6x^2-18x-8}{(3x^2-4)^2} \end{aligned} So we found that $\dfrac{d}{dx}\left(\dfrac{2x+3}{3x^2-4}\right)=\dfrac{-6x^2-18x-8}{(3x^2-4)^2}$. Now let's plug $x= {-1}$ : $\begin{aligned} &\phantom{=}\dfrac{-6( {-1})^2-18( {-1})-8}{(3( {-1})^2-4)^2} \\\\ &=\dfrac{-6+18-8}{(-1)^2} \\\\ &=4 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(\dfrac{2x+3}{3x^2-4}\right)$ at $x=-1$ is $4$.